Problem: Simplify and expand the following expression: $ \dfrac{3y + 4}{2y - 2}-\dfrac{y}{3y - 6} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2y - 2)(3y - 6)$ Multiply the first term by $\dfrac{3y - 6}{3y - 6}$ $ \begin{align*} \dfrac{3y + 4}{2y - 2} \times \dfrac{3y - 6}{3y - 6} & = \dfrac{(3y + 4)(3y - 6)}{(2y - 2)(3y - 6)} \\ & = \dfrac{9y^2 - 6y - 24}{(2y - 2)(3y - 6)}\end{align*} $ Multiply the second term by $\dfrac{2y - 2}{2y - 2}$ $ \begin{align*} \dfrac{y}{3y - 6} \times \dfrac{2y - 2}{2y - 2} & = \dfrac{(y)(2y - 2)}{(3y - 6)(2y - 2)} \\ & = \dfrac{2y^2 - 2y}{(3y - 6)(2y - 2)}\end{align*} $ Now we have: $ = \dfrac{9y^2 - 6y - 24}{(2y - 2)(3y - 6)} - \dfrac{2y^2 - 2y}{(3y - 6)(2y - 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{9y^2 - 6y - 24 - (2y^2 - 2y)}{(2y - 2)(3y - 6)} $ $ = \dfrac{9y^2 - 6y - 24 - 2y^2 + 2y}{(2y - 2)(3y - 6)} $ $ = \dfrac{7y^2 - 4y - 24}{(2y - 2)(3y - 6)}$ Expand the denominator: $ = \dfrac{7y^2 - 4y - 24}{6y^2 - 18y + 12}$